#!/usr/bin/env python
# -*- coding: utf-8 -*-
# __author__:vincentlc
# time: 16/5/30 : 16:59
'''
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,
              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
'''


# Definition for a binary tree node.
class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        self.val = sum
        return  self.hasPathSum2(root,0, sum)

    def hasPathSum2(self, root, sum, val):
        flag1 = False
        flag2 = False
        flag = False
        if root != None:
            sum += root.val
            if root.left != None:
                flag1 = self.hasPathSum2(root.left, sum, val)

            if root.right != None:
                flag2 = self.hasPathSum2(root.right, sum, val)

            if root.left == None and root.right == None and sum == val:
                flag = True

            return flag1 or flag2 or flag
        else:
            return False


N_a = TreeNode(1)
N_a.left = TreeNode(2)
cls_a = Solution()
result = cls_a.hasPathSum(N_a,3)
print(result)
